Question: The equation of a circle $C$ is $x^2+y^2+8x+12y+16 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+8x) + (y^2+12y) = -16$ $(x^2+8x+16) + (y^2+12y+36) = -16 + 16 + 36$ $(x+4)^{2} + (y+6)^{2} = 36 = 6^2$ Thus, $(h, k) = (-4, -6)$ and $r = 6$.